Matrix Commutativity - Integration Specifications and Data in question
1.We have a skew symmetric matrix 
$ A(t)_{3\times 3}=  \begin{bmatrix}\,0&\!-a_3(t)&\,\,a_2(t)\\ \,\,a_3(t)&0&\!-a_1(t)\\-a_2(t)&\,\,a_1(t)&\,0\end{bmatrix} \tag 1$ and
$B(t)_{3 \times 3}=   -\int_{0}^{t} A(s)\ ds \tag 2$  
$a_1(t),a_2(t),a_3(t)$ are general non matrix functions.  Just used   here to show the form of $A(t)$
Question


*

*Is it possible to prove $B(t)*B(t)^{'}=B(t)^{'}*B(t)$  (commutative)? if not why?
NB:  * means multiplication and  $B'(t) =\frac{\text{d}B(t) }{\text{d}t} $
 A: It is sufficient to find any two non-commuting (skew-symmetric) matrix $P,Q$.  Let $p_{i},q_{i}$ denote the entries of $P,Q$ respectively.  If we then define
$$
B(t) = 
\pmatrix{0&-p_{3}\;t^{q_3/p_3} & p_2\;t^{q_2/p_2}\\
p_{3}\;t^{q_3/p_3} & 0 & -p_{1}\;t^{q_1/p_1}\\
-p_2\;t^{q_2/p_2}&p_{1}\;t^{q_1/p_1}&0}
$$
Then $B'(1)B(1) \neq B(1)B'(1)$.  A similar trick using $\sin(t)$ or $e^t$ gives you such a $B(t)$ where the function is necessarily smooth at $0$.
A: In general the answer is negative as $B(t)$ and $B'(t)$ satisfy
$$B'(t)B(t)+B(t)B'(t)=\frac{d}{dt}\left(B^2(t)\right). $$
You can translate the commutativity condition  
$$B'(t)B(t)=B(t)B'(t)~~(*)$$ 
into a system of equations for the elements of $B(t)$ and $B'(t)$. Explicitly, if
$$
B(t) = 
\pmatrix{ 0 &\alpha(t) & \beta(t) \\
-\alpha(t) & 0 & \gamma(t)\\
-\beta(t) & -\gamma(t) & 0}
$$
then commutativity $(*)$ is equivalent to $\alpha(t)\gamma'(t)=\alpha'(t)\gamma(t),$
$\alpha(t)\beta'(t)=\alpha'(t)\beta(t),$ and $\beta(t)\gamma'(t)=\beta'(t)\gamma(t)$.
The triple  
$$\alpha(t):=t,~~\beta(t):=-t,~~\gamma(t)=t^3$$
gives you a counterexample, for all $t$. (you can quickly derive the original skew symmetric matrix $A(t)$ for such a triple). 
